Decomposing real line as a union of a nullset and a set of first category $\Bbb R$ can be written of the form $A\cup B$ such that $A$ is of measure zero and $B$ is of the first category!
can anybody prove this?
I guess $A$ must be an $G_{\delta}$ set which is dense in $\Bbb R$ and obviously $B=\Bbb R-A$.
 A: Enumerate the rational numbers as a sequence $\{ r_n;\; n\in\mathbb N\}$. For each $n\in\mathbb N$ and all $j\in\mathbb N$, set $$I_{n,j}:=\left] r_n-\frac{1}{j}\, 2^{-n}, r_n+\frac{1}j\, 2^{-n}\right[\, .$$ 
Then define 
$$O_j:=\bigcup_{n\in\mathbb N} I_{n,j}\, , $$
and 
$$ A:=\bigcap_{j\in\mathbb N} O_j\, .$$
Each $O_j$ is an open set containing all the $r_n$, so $A$ is a $G_\delta$ set containing all the $r_n$ and hence a dense $G_\delta$ set. Moreover, denoting by $\mu$ the Lebesgue measure on $\mathbb R$, we have
$$\mu(A)\leq\mu(O_j)\leq\sum_{n=1}^\infty \mu(I_{n,j})=\sum_{n=1}^\infty \frac{2}j\, 2^{-n}=\frac{2}j $$
for all $j\in\mathbb N$, so that $\mu(A)=0$.
A: Here is another solution along the same lines:
Let $\{r_n:n\in\mathbb{N}\}$ be an enumeration of the rational numbers.
Let $I_{nj}$ denote the open interval centered at $r_n$ of length $2^{-n-j}$.
Each set $U_j=\bigcup_nI_{nj}$ is open and contained the rationals; hence $U_j$ is an open dense set in $\mathbb{R}$. By Baire's theorem $B=\bigcap_j U_j$ is dense in $\mathbb{R}$ and $A=\mathbb{R}\setminus B=\bigcup_j(\mathbb{R}-U_j)$ is a set of first category: $\emptyset=\mathbb{R}\setminus \overline{U_j}=\operatorname{Int}(\mathbb{R}\setminus U_j)$ for each $j$.
Denoting by $\lambda$ the Lebesgue measure on $\mathbb{R}$ we have
$$\lambda(B)\leq \lambda(U_j)\leq \sum_n\lambda(I_{nj})=2^{-j}\qquad\forall j\in\mathbb{N}$$
This implies that $\lambda(B)=0$.

Another, perhaps more interesting, decomposition can be obtained through use of a little of number theory motivated by a nice Wikipedia article on Liouville numbers
First some generalities:

*

*Recall that $x\in\mathbb{R}$ is an  algebraic number if there is a polynomial $p(x)$ with coefficients in $\mathbb{Z}$ such that $p(x)=0$.

*If $x$ is algebraic, the smallest degree of all polynomials $p$ for which $p(x)=0$ is called the order of $x$. It follows that rational numbers are algebraic and that $x$ is an algebraic number of order $1$ iff $x$ is rational.

*Since the collection of polynomials with coefficients in $\mathbb{Z}$ is countable, then the collection of algebraic numbers is countable.

*Numbers that are not albegraic are called transcendental. It follows that the collection of transcendental numbers in uncountable.

Definition: A Liouville number $x$ is an irrational number such that for for any $n\in\mathbb{N}$, there are $p,q\in\mathbb{Z}$ such that $q>1$, and
$$\Big|z-\frac{p}{q}\Big|<\frac{1}{q^n}$$
We have now all the ingredients to construing the decomposition. Let $E$ denote the set of Liouville numbers. Then
$$ E=\big(\mathbb{R}\setminus\mathbb{Q}\big)\cap\bigcap_{n\in\mathbb{N}}\left(\bigcup^\infty_{q=2}\bigcup_{p\in\mathbb{Z}}\big(\frac{p}{q}-\frac{1}{q^n},\frac{p}{q}+\frac{1}{q^n}\big)\right)
$$
For each $n\in\mathbb{N}$, the set $U_n=\bigcup^\infty_{q=2}\bigcup_{p\in\mathbb{Z}}\big(\frac{p}{q}-\frac{1}{q^n},\frac{p}{q}+\frac{1}{q^n}\big)$ is an open set that containes the rational numbers, thus $U_n$ is open and dense in $\mathbb{R}$, and by Baire's theorem, $\bigcap_nU_n$ is dense in $\mathbb{R}$. It follows that $\bigcup_n\mathbb{R}\setminus U_n$ is of first category and so,
$$\mathbb{R}\setminus E=\mathbb{Q}\cup\bigcup_n\mathbb{R}\setminus U_n$$
is of first category.
For each integer $q\geq 2$, let $U_{qn}=\bigcup_{p\in\mathbb{Z}}\big(\frac{p}{q}-\frac{1}{q^n},\frac{p}{q}+\frac{1}{q^n}\big)$.
Since $E\subset U_n$, for any $m\in\mathbb{N}$,
$$
\begin{align}
E\cap(-m,m)&\subset U_m\cap(-m,m)=\bigcup_{q\geq2}G_{qn}\cap(-m,m)\\
&\subset\bigcup_{q\geq2}\bigcup^{mq}_{-mq}\big(\frac{p}{q}-\frac{1}{q^n},\frac{p}{q}+\frac{1}{q^n}\big)
\end{align}
$$
Consequently, for any $n>2$
$$
\begin{align}
\lambda(E\cap(-m,m))&\leq \sum_{q\geq2}\sum^{mq}_{p=-mq}\frac{2}{q^n}=\sum_{q\geq2}(2mq+1)\frac{2}{q^n}\\
&\leq (4m+1)\sum_{q\geq2}\frac{1}{q^{n-1}}\leq (4m+1)\int^\infty_1\frac{dx}{x^{n-1}}=\frac{4m+1}{n-2}
\end{align}
$$
whence we conclude that $\lambda(E\cap(-m,m))=0$ and so, $\lambda(E)=0$.

What makes this construction interesting is that it shows that $E$ is of second category (so it is a rather fat set in the topological set) and yet it has measure $0$. This is to be compared with the following well known result in analytic number theory:
The following is a well known result in analytic number theory:
Theorem: Every Liouville number is transcendental.
Thus, mostly all transcendental numbers are not of the Liouville type!
